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tiL12214_appd_633-642.indd Page 639 09/10/10 8:36 AM user-f463
tiL12214_appd_633-642.indd Page 639 09/10/10 8:36 AM user-f463 Volume/207/MHDQ243/tiL12214_disk1of1/0073512214/tiL12214_pagefile

( day length mean solar ) Example 17.4, p. 444
__
sidereal days = year length

day length sidereal Read the percentage associated with the range of each mineral from
Figure 17.13, and calculate the difference to determine the percentage
Convert time of sidereal day to hours.
of each mineral at each boundary of the range in composition.
_ _ _
1 h
1 min
1 h
day length sidereal = 23 h + 56 min ( ) + 4 s ( ) ( ) Basalt (high silica boundary)

60 min 60 min 60 s
Pyroxene 100% − 50% = 50%
4 _ _ _
56 _
h _
h
min
= 23 h + min ( ) + s ( ) ( ) Calcium-rich plagioclase 50% − 0% = 50%

60 min 60 2 min s
= 23 h + 0.9333 h + 0.0011 h Basalt (low silica boundary)
= 23.9344 h Pyroxene/Olivine 100% − 0% = 100%
Calcium-rich plagioclase 0% − 0% = 70%
(
__
= year length day length mean solar ) Range

day length sidereal
Pyroxene 50% to 100%
h _
_)

24 Calcium-rich plagioclase 0% to 50%
__
msd

= 365.25636 msd

( 23.9344 h

day sidereal
CHAPTER 18
h _
_)

_ _
24
msd

= 365.25636 ( ) msd Example 18.2, p. 458

23.9344 h

( ρ crust

mantle )

day sidereal z crust = 35 km h = z crust − z crust _
( ρ
g
= 365.25636 (1.00274) day sidereal _
ρ = 3.3 _
g

= 367.247 day sidereal mantle c m 3 ( 3.3 )
2.7

3
_
cm
g
_
g
ρ crust = 2.7 = 35.0 km − 35.0 km _

cm 3 c m
3
CHAPTER 17 h = ?
g
_

Example 17.2, p. 440 2.7 _ _ 3
c m

The volume of the mineral can be calculated by the formula for a = 35.0 km − 35.0 ( ) cm _
g
3.3

rectangular solid. Determine the density of the mineral specimen by c m 3
dividing its mass by its volume calculated from the dimensions of = 35.0 km − 35.0(0.82) km
the specimen. Then divide this by the density of water to determine
specific gravity. = 35.0 km − 28.7 km
= 6.3 km
L = 1.82 cm
W = 1.18 cm From Example 18.1:
H = 1.16 cm
h oceanic = 0.6 km
m = 6.39 g
difference = h continental − h oceanic
g
_

ρ water = 1 = 6.3 km − 0.6 km
cm 3
= 5.7 km
SG = ?
_ m _ Example 18.4, p. 467
ρ mineral
SG = and ρ mineral = and V = L × W × H

ρ water V
As Figure 18.18 shows, earthquake foci delineate the upper surface of
__ the subducting plate, so the depth and distance of an earthquake from
m

__
(L × W × H)

∴ SG = the trench can be used to determine the slope of the subducting plate.
ρ water
Calculate the slope based on each of the distance/depth pairs, then
6.39 g
___ average the speeds. The y value is negative because it is a depth.

(1.82 cm)(1.18 cm)(1.16 cm)

= ___
g
_
1 Data Pair 1

Δy
cm 3 _
Δx = 8.0 km slope =

Δx
g
__ __ Δy = −1.9 km
6.39

(1.82)(1.16)(1.18) (cm)(cm)(cm)
= __ __ slope = ? Convert the y value from km to m:

g
1 _ 3

1 × 1 0 m

cm 3 −1.9 km ( _ )

1 km
6.39
3
= _ −1.9 × 1 0 m
2.49
3
−1.9 × 1 0 m
= 2.56 slope = __

8.0 km
m _
= −240
km
D-7 APPENDIX D Solutions for Follow-Up Example Exercises 639